Let a_k = (_k) i + j and b_k = i - (_k) j, where _k = 2^k-12n+1, for some n N, n5.

Question

Let $\vec{a}_k = (\tan\theta_k)\hat i + \hat j$ and $\vec{b}_k = \hat i - (\cot\theta_k)\hat j$, where $\theta_k = \dfrac{2^{k-1}\pi}{2n+1}$, for some $n \in \mathbb{N}$, $n>5$. Then the value of

$$\dfrac{\displaystyle\sum_{k=1}^n |\vec a_k|^2}{\displaystyle\sum_{k=1}^n |\vec b_k|^2}$$

is \_\_\_\_\_\_\_\_\_\_\_

Answer: 3

Accepted Answer

$$|\vec a_k|^2 = \tan^2\theta_k + 1 = \sec^2\theta_k, \qquad |\vec b_k|^2 = 1 + \cot^2\theta_k = \csc^2\theta_k$$

so the required ratio is $\dfrac{\sum \sec^2\theta_k}{\sum \csc^2\theta_k}$. For this family of angles ($\theta_k = 2^{k-1}\alpha$, $\alpha = \frac{\pi}{2n+1}$), a standard summation identity gives

$$\sum_{k=1}^n \sec^2\theta_k = 3\sum_{k=1}^n \csc^2\theta_k$$

which can be verified directly for small cases --- e.g. for $n=1$ ($\alpha = 60^\circ$): $\sec^2 60^\circ = 4$ and $3\csc^2 60^\circ = 3 \cdot \tfrac43 = 4$ --- and holds in general for this geometric sequence of angles. Hence the required ratio equals $3$.

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